How can I determine how many characters are in a string?
The problem I face is the stipulation that I cannot use the <string.h> library.
Here is what I'm trying to solve:
#include <assert.h>
#include <stdio.h>
#define MAX_LENGTH 1024
int string_length(char *string);
int main(int argc, char *argv[]) {
assert(string_length("") == 0);
assert(string_length("!") == 1);
assert(string_length("Hello, world!") == 13);
assert(string_length("17... seventeen.\n") == 17);
printf("All tests passed. You are awesome!\n");
return 0;
}
int string_length(char *string) {
// Add code here
return 0;
}
All the strings in C end with a special char \0 so you just need to check how many chars there are before that char.
int string_length(char *string) {
int chars = 0;
while(*string != '\0'){
chars++;
string++;
}
return chars;
}
Related
I'm trying to creat a program that removes all occurences from a character that I choose from a certain string and also returs the number of characters that were removed.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define DIMV 10
int eliminar(char texto[], char ch, char novoTexto[]){
int a=0,i;
for(i=0; texto[i] != '\0';i++){
texto[i] = texto[i+a];
novoTexto[i]=texto[i];
if(novoTexto[i]==ch){
novoTexto[i] = '\0';
a++;
}
}
return a;
}
int main()
{
char frase[]="Uma arara torta!";
char res[50];
printf("%s\n",frase);
printf("%d\n",eliminar(frase,'r',res));
printf("%s\n",res);
return 0;
}
When I run the program, it returns:
Uma arara torta!
3
Uma a
What I wanted to return is:
Uma arara torta!
3
Uma aaa tota!
It's a silly question but I can't find the mistake that I made. Thank you!
Putting '\0' means to terminate the string there.
You have to skip adding characters to eliminate.
Try this:
int eliminar(char texto[], char ch, char novoTexto[]){
int a=0,i,j=0;
for(i=0; texto[i] != '\0';i++){
if(texto[i]==ch){
a++;
}else{
novoTexto[j]=texto[i];
j++;
}
}
novoTexto[j]='\0';
return a;
}
Putting '\0' means to terminate the string there.
check this:
int eliminator(char * texto, char ch, char * novoTexto) {
int i=0;
while(*texto != '\0') {
if (*texto == ch) {
texto++;
++i;
} else {
*novoTexto++ = *texto++;
}
}
*novoTexto = '\0';
return i;
}
int main(int arg, char **argv) {
char frase[] = "Uma arara torta!";
char res[50];
printf("%s\n",frase);
printf("%d\n",eliminator(frase, 'r', res));
printf("%s\n",res);
}
Here's my code:
#include <stdio.h>
#include <string.h>
char input_buffer[1000];
void get_substring(){
int i;
int length;
printf("Please enter a string:\n");
scanf("%[^\n]s", input_buffer);
printf("Index of first character of substring:\n");
scanf("%d", &i);
printf("Length of substring:\n");
scanf("%d", &length);
printf("Substring is %.*s ", length, input_buffer + i);
}
int main(void) {
// your code goes here
//get_substring(0,4);
get_substring();
return 0;
}
That's my current code, I want to return a pointer of the input, instead of just displaying the substring. Sorry for the confusion everyone.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* getSubstring(char* str,size_t start, size_t length)
{
// determine that we are not out of bounds
if(start + length > strlen(str))
return NULL;
// reserve enough space for the substring
char *subString = malloc(sizeof(char) * length);
// copy data from source string to the destination by incremting the
// position as much as start is giving us
strncpy(subString, str + start, length);
// return the string
return subString;
}
int main(int argc, char* argv[])
{
char *str = "Hallo Welt!";
char *subStr = getSubstring(str,0,20);
if(subStr != NULL)
{
printf("%s\n",subStr);
free(subStr);
}
}
This solution should give you a hint how you would start with such a problem.
My code is giving me a segmentation fault. I'm 99% sure the fault is stemming from my lousy code construction.
#include <stdio.h>
#include <assert.h>
#include <string.h>
int decToBit(unsigned int I, char *str){
str = "";
int currentVal = I;
do{
if(I%2 == 0)
strcat(str,"0");
else
strcat(str,"1");
} while(currentVal > 0);
return(0);
}
You need to make sure that there is enough space in str to add the extra characters:
char myStr[200];
myStr[0] = '\0'; // make sure you start with a "zero length" string.
strcpy(myStr, str);
and then use myStr where you were using str.
As it is, the statement
str="";
points str to a const char* - that is a string you can read but not write.
Incidentally the call signature for main is
int main(int argc, char *argv[])
in other words, you need a pointer to a pointer to char. If I am not mistaken, you would like to do the following (a bit of mind reading here):
Every odd argument gets a 1 added; every even argument gets a 0 added.
If my mind reading trick worked, then you might want to try this:
#include <stdio.h>
#include <string.h>
int main(int argc, char * argv[]) {
char temp[200];
temp[0] = '\0';
int ii;
for(ii = 0; ii < argc; ii++) {
strncpy(temp, argv[ii], 200); // safe copy
if(ii%2==0) {
strcat(temp, "0");
}
else {
strcat(temp, "1");
}
printf("%s\n", temp);
}
}
edit just realized you edited the question and now your purpose is much clearer.
Modified your function a bit:
int decToBit(unsigned int I, char *str){
str[0] = '\0';
char *digit;
do
{
digit = "1";
if ( I%2 == 0) digit = "0";
strcat(str, digit);
I>>=1;
} while (I != 0);
return(0);
}
It seems to work...
In do-while loop you should increment the value of currentVal. Otherwise it will be an infinity loop and you will end up with Segmentation fault.
Initialize str[0] properly.
Divide I by 2 each loop.
But then the string will be in a little endian order. Doubt that was intended?
int decToBit(unsigned int I, char *str) {
str[0] = '\0';
do {
if (I%2 == 0)
strcat(str,"0");
else
strcat(str,"1");
I /= 2;
} while(I > 0);
return(0);
}
// call example
char buf[sizeof(unsigned)*CHAR_BIT + 1];
decToBit(1234567u, buf);
#include <stdio.h>
#include <string.h>
#include <limits.h>
#include <assert.h>
char *decToBit(unsigned int I, char *str){
int bit_size = CHAR_BIT * sizeof(I);
str += bit_size;
*str = 0;
do{
*--str = "01"[I & 1];
}while(I>>=1);
return str;
}
int main(){
char bits[33];
printf("%s\n", decToBit(0, bits));
printf("%s\n", decToBit(-1, bits));
printf("%s\n", decToBit(5, bits));
return 0;
}
I'm new to C and trying to create a function that check a string and returns the last character.
I get the function to print the correct letter, but I cant figure out how to return it :/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char last_chr(char *c);
int main (int argc, const char * argv[]) {
char *text[15];
strcpy(text, "*find:last;char#");
last_chr(text); //debugging
//printf("last char: %c", last_chr(text)); //not working
return 0;
}
char last_chr(char *c) {
char *endchr;
char result;
int pos = strlen(c)-1;
endchr = c[pos];
//sprintf(result,"%s",endchr); //"EXEC_BAD_ACCESS"
putchar(endchr); //prints #
//putc(endchr, result); //"EXEC_BAD_ACCESS"
//printf(endchr); //"EXEC_BAD_ACCESS"
return result;
}
You don't assign result. You probably mean
result = c[pos];
instead of endchr = c[pos];
endchr is a character-pointer instead of a character.
#include <stdio.h>
#include <string.h>
char last_chr(char *c);
int main (int argc, const char * argv[]) {
char text[32];//char *text[15]
strcpy(text, "*find:last;char#");//length is 17
printf("last char: %c", last_chr(text));//#
return 0;
}
char last_chr(char *c) {
if(c == NULL || *c == '\0') return 0;
return c[strlen(c)-1];
}
Is there a way to print a string of fixed size in reverse without using pointers?
#include<stdio.h>
main()
{
char buffer[10];
scanf("%s", buffer);
// need to print buffer in reverse without using pointers??
}
A lovely K&R function to reverse your string in-place before printing it, perhaps?
#include <stdio.h>
#include <string.h>
void strrev(char *s) {
int tmp, i, j;
for (i = 0, j = strlen(s) - 1; i < j; i++, j--) {
tmp = s[i];
s[i] = s[j];
s[j] = tmp;
}
}
int main(int argc, const char *argv[]) {
char buffer[10];
scanf("%s", buffer);
strrev(buffer);
printf("%s\n", buffer);
return 0;
}
#include<stdio.h>
main()
{
char buffer[10];
int n = scanf("%s", buffer);
// print the number of chars written to buffer
if (n != EOF) {
int len = strlen(buffer);
if (len <= 10) {
int i;
for (i = len - 1; i >= 0; i--)
printf("%c", buffer[i]);
}
}
}
Since [] is just syntactic sugar for pointers, here's a version that works completely without pointers, arrays or anything else, just one single int. You didn't say that the string has to be stored somehow. :) (Note that I use fgetc instead of a buffer and scanf).
[jkramer/sgi5k:.../c]# cat rev.c
#include <stdio.h>
#include <stdlib.h>
void read_print();
int main(void) {
fputs("Enter your string, yo! ", stdout);
read_print();
fputs("\nDone!\n", stdout);
return EXIT_SUCCESS;
}
void read_print() {
int c = fgetc(stdin);
if(c != EOF && c != '\n') {
read_print();
fputc(c, stdout);
}
}
[jkramer/sgi5k:.../c]# gcc -o rev rev.c -Wall -W -Os
[jkramer/sgi5k:.../c]# ./rev
Enter your string, yo! foobar
raboof
Done!
Here's a recursive way of doing it; technically, this is using a pointer, but I wouldn't go into language-lawyer mode with such simple tasks.
#include <stdio.h>
/* If you want it printed forward, or backward, or think of another way.. */
typedef enum {
FRONT = 1,
BACK,
} direction;
/* Technically still using a pointer...don't nitpick. */
void echo_string(char buffer[], size_t buflen, direction from)
{
/* An index into the buffer to echo, which will preserve
* its value across subsequent recursive calls.
*/
static size_t index = 0;
/* According to the specified direction, print from the front
* or the back of the buffer. Advance the index (a misnomer, I guess).
*/
if(from == FRONT) {
printf("%c", buffer[index++]);
}
else {
printf("%c", buffer[buflen - ++index]);
}
/* Are there any more characters to echo? Yes? Awesome! */
if(index != buflen) {
echo_string(buffer, buflen, from);
}
}
int main(int argc, char **argv)
{
char buffer[10];
scanf("%s", buffer);
/* Better strlen() than sizeof() here,
* but BEWARE! scanf() is DANGEROUS!
*/
echo_string(buffer, strlen(buffer), BACK);
return(0);
}
reverse(char c[], int len)
{
if( ! (len / 2))
return;
char t = c[0];
c[0] = c[len--];
c[len] = t;
reverse(c, len-1);
}
The error(s) is left as an exercise to the student.
As caf pointed out, we're still using pointers..!
Here's an other way to solve the problem (of reversing a string).
This code snippet (and probably most others) don't respect stuff like utf8. I think signines post demonstrating the K&R way was quite close to mine (:D) so I adapted mine to fit that example (and corrected some things..)
#include <stdio.h>
#include <string.h>
void strrev(char *s) {
size_t len = strlen(s) + 1;
size_t i, j;
for(i = 0; i < len / 2; i++) {
j = len-1 - i-1;
char tmp = s[j];
s[j] = s[i];
s[i] = tmp;
}
}
int main(int argc, const char *argv[]) {
char buffer[10];
scanf("%s", buffer); // Look out for an overflow ;)
strrev(buffer);
puts(buffer);
return(0);
}
You can use strrev to reverse a string.
#include <stdio.h>
#include <string.h>
main()
{
char buffer[10];
scanf("%s", buffer);
strrev(buffer);
printf("%s", buffer);
}
void outstrreverse(const char s[])
{
size_t l=strlen(s);
while( l && s!=&s[--l] )
putchar(s[l]);
if(s[0])
putchar(s[0]);
}
Because of the relationship between C strings, arrays, and pointers the exercise is rather shotty IMHO - the most idiomatic description of a "String" in C is represented by the char*, which is not an array. Your (the OPs) title and post differ in their definitions between string and char[fixed length].
The OP should read and understand this FAQ entry, and between that and the posts here: easily figure out a solution—as well as defend it to the teacher/judge if need be.
I'll comment on this: never use scanf("%s", buffer) to populate a fixed length string. If you must use scanf() to do it, please use a field width specifier: e.g. scanf("%9s", buffer); if buffer is an [10], you want a specifier of 9 because of how scanf fills the buffer: otherwise you must beware the dragons! You could also scanf by character and evade the issue with a loops bounds, but that would likely be less efficient.
#include <stdio.h>
#include <conio.h>
void reverse(char a[], int s, int sc );
void reverse(char a[], int s, int sc ){
if ((sc-s)<(s-1))
{
a[sc-s]^=a[s-1];
a[s-1]^=a[sc-s];
a[sc-s]^=a[s-1];
reverse (a, s-1, sc) ;
}
}
void main (){
char a[]="ABCDEFG";
reverse(a, 7, 7);
printf("%d",a);
getch(); //i just use it to freeze the screen
}